Đa thức một biến - Bài tập nâng cao

Home Toán lớp 7 - Sách KNTT Bài 25: Đa thức một biếnBài tập nâng cao

{"common":{"save":0,"post_id":"5808","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"8490","post_id":"5808","mon_id":"1158532","chapter_id":"1158660","question":"Xác \u0111\u1ecbnh b\u1eadc c\u1ee7a các \u0111a th\u1ee9c sau:<\/p>$\\dfrac{2}{7}x-7+6x^3$<\/span>  có b\u1eadc là <input class=\"tim_x input_number\" tpl=\"dien_so\" type=\"text\"\/><\/p>$0,5x^{2022}+2022+x-\\dfrac{1}{2}x^{2022}$<\/span>  có b\u1eadc là <input class=\"tim_x input_number\" tpl=\"dien_so\" type=\"text\"\/><\/p>$100y^5-0,1y^7+6x^3$<\/span>  có b\u1eadc là <input class=\"tim_x input_number\" tpl=\"dien_so\" type=\"text\"\/><\/p>","options":"","correct":["[\"3\",\"1\",\"7\"]"],"level":"3","hint":"Trong m\u1ed9t \u0111a th\u1ee9c thu g\u1ecdn và khác \u0111a th\u1ee9c không:  + B\u1eadc c\u1ee7a \u0111a th\u1ee9c là b\u1eadc c\u1ee7a h\u1ea1ng t\u1eed có b\u1eadc cao nh\u1ea5t<\/span> + H\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c là h\u1ec7 s\u1ed1 c\u1ee7a h\u1ea1ng t\u1eed có b\u1eadc cao nh\u1ea5t + H\u1ec7 s\u1ed1 t\u1ef1 do là h\u1ec7 s\u1ed1 c\u1ee7a h\u1ea1ng t\u1eed có b\u1eadc 0<\/p>","answer":"$\\dfrac{2}{7}x-7+6x^3$<\/span>  có b\u1eadc là 3 (vì h\u1ea1ng t\u1eed có b\u1eadc cao nh\u1ea5t là $6x^3$<\/span>)<\/p>$0,5x^{2022}+2022+x-\\dfrac{1}{2}x^{2022}=2022+x$<\/span>  có b\u1eadc 1 (thu g\u1ecdn \u0111a th\u1ee9c thì h\u1ea1ng t\u1eed có b\u1eadc cao nh\u1ea5t là x)<\/p>$100y^5-0,1y^7+6x^3$<\/span>  có b\u1eadc là  7 (vì hàng t\u1eed có b\u1eadc cao nh\u1ea5t là $-0,1x^7$<\/span>)<\/p>V\u1eady các s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t là 3  ;  1  ; 7<\/span><\/p>","type":"blank","extra_type":"","time":"0","user_id":"131","test":"2","date":"2022-05-31 10:06:22"},{"id":"8492","post_id":"5808","mon_id":"1158532","chapter_id":"1158660","question":"\u0110a th\u1ee9c nào sau \u0111ây có h\u1ec7 s\u1ed1 t\u1ef1 do b\u1eb1ng 2022 ? <\/p>","options":["A.<\/strong> $x^4-2022x$<\/span>","B. <\/strong>$-2022+x^2-0,5x^4$<\/span>","C. <\/strong>$x^{2022} +1$<\/span>","D. <\/strong>$2023+x^4-\\dfrac{x^7}{2}-1099^0$<\/span>"],"correct":"4","level":"3","hint":"Trong m\u1ed9t \u0111a th\u1ee9c thu g\u1ecdn và khác \u0111a th\u1ee9c không:  + B\u1eadc c\u1ee7a \u0111a th\u1ee9c là b\u1eadc c\u1ee7a h\u1ea1ng t\u1eed có b\u1eadc cao nh\u1ea5t + H\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c là h\u1ec7 s\u1ed1 c\u1ee7a h\u1ea1ng t\u1eed có b\u1eadc cao nh\u1ea5t + <\/span>H\u1ec7 s\u1ed1 t\u1ef1 do là h\u1ec7 s\u1ed1 c\u1ee7a h\u1ea1ng t\u1eed có b\u1eadc 0<\/span><\/p>(H\u1ea1ng t\u1eed có b\u1eadc 0 là các s\u1ed1 th\u1ef1c khác 0)<\/p>","answer":"A.<\/strong> $x^4-2022x$<\/span>  không có h\u1ec7 s\u1ed1 t\u1ef1 do <\/span>(vì không có h\u1ea1ng t\u1eed nào có b\u1eadc 0)<\/p>B. <\/strong>$-2022+x^2-0,5x^4$<\/span> có h\u1ec7 s\u1ed1 t\u1ef1 do là $-2022$<\/span> <\/span>(vì h\u1ea1ng t\u1eed có b\u1eadc 0 là $-2022$<\/span><\/span>)<\/p>C. <\/strong>$x^{2022} +1$<\/span> có h\u1ec7 s\u1ed1 t\u1ef1 do là 1  <\/span>(vì h\u1ea1ng t\u1eed có b\u1eadc 0 là 1)<\/p>D. <\/strong>$2023+x^4-\\dfrac{x^7}{2}-1099^0$<\/span> = $x^4-\\dfrac{x^7}{2}+2022$<\/span> có h\u1ec7 s\u1ed1 t\u1ef1 do là 2022<\/p>\u0110áp án \u0111úng là D. <\/strong>$2023+x^4-\\dfrac{x^7}{2}-1099^0$<\/span><\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"1","date":"2022-05-31 10:15:25","option_type":"math","len":0},{"id":"8494","post_id":"5808","mon_id":"1158532","chapter_id":"1158660","question":"Xác \u0111\u1ecbnh b\u1eadc c\u1ee7a \u0111a th\u1ee9c sau:<\/p>$4x^{2020}+0,3x^2-\\dfrac{5}{2}x^7-2^2x^{2020}+3000^0$<\/span><\/p>","options":["2020","7","2","0"],"correct":"2","level":"3","hint":"Rút g\u1ecdn \u0111a th\u1ee9c trên<\/p>","answer":"Ta có: $4x^{2020}+0,3x^2-\\dfrac{5}{2}x^7-2^2x^{2020}+3000^0$<\/span><\/p>  = ($4x^{2020}-2^2x^{2020}$<\/span>)$-\\dfrac{5}{2}x^7$<\/span>$+0,3x^2$<\/span>$+3000^0$<\/span><\/p>  = $(4-2^2).x^{2020}$<\/span>$-\\dfrac{5}{2}x^7$<\/span>$+0,3x^2$<\/span>$+1$<\/span><\/p>  = $-\\dfrac{5}{2}x^7$<\/span>$+0,3x^2$<\/span>$+1$<\/span><\/p>Trong d\u1ea1ng thu g\u1ecdn c\u1ee7a \u0111a th\u1ee9c trên, h\u1ea1ng t\u1eed có b\u1eadc cao nh\u1ea5t là $-\\dfrac{5}{2}x^7$<\/span> (có b\u1eadc là 7) nên b\u1eadc c\u1ee7a \u0111a th\u1ee9c là 7<\/p>V\u1eady \u0111áp án \u0111úng là 7.<\/span><\/span><\/p>","type":"choose","extra_type":"star","time":"0","user_id":"131","test":"3","date":"2022-05-31 10:23:20","option_type":"txt","len":0}]}